# What does it mean when two Groups are isomorphic?

I'm not asking for the formal definition I know it. An isomorphism is a bijective homomorphism. In my book it's indicated many times when two groups are isomorphic, and I don't understand what's the reason for that. What can we "do" when we know that 2 groups are isomorphic? What does it really mean when 2 groups are isomorphic?

• If you care only about the group structure and not about what the elements are, then the two are exactly the same.
– Ian
Nov 27, 2015 at 18:54
• If this comment doesn't help you then just ignore it. A pair of isomorphic groups are, from an intuitive standpoint, a bit like a pair of isomorphic graphs: they may have different names, and they may even look different, but they're actually the same with regards to how the individual elements interact with each other. Nov 27, 2015 at 19:26
• @WillR: I really like this analogy. I'm stealing it. Thanks!
– user98602
Nov 28, 2015 at 3:54
• @MikeMiller: if you like to fiddle around with Cayley tables, you'll find that sometimes two Cayley tables can look quite different until you decide to swap two well-chosen columns; in my mind this is just like moving around the vertices of a graph: two graphs can look different, but if you decide to move just the right vertices they immediately look very similar. The downside to the analogy is that I can't find any actual general connection; for example, the cycle graph does not uniquely represent the group. Nov 28, 2015 at 4:16
• "when?", 're they time dependent?
– DVD
Dec 28, 2015 at 22:35

It means they are exactly the same except for the names of the elements and the name of the binary operation. An isomorphism between groups is a function that renames all of the elements. (Hence, it is bijective... each element in the first group gets renamed to be exactly one element in the second group.)

The reason we care is that if you are only concerned with the group structure, then the names of the elements or the symbol you use for the binary operation aren't terribly important. Thus, if you know two groups are isomorphic everything about them, in a group theoretic sense, is the same. This is nice since if you can show a group you encounter is isomorphic to a group you already know about, then you get any group-theoretic property of your new group for free.

• Wow, it seems a very powerfull statement to say that two groups are isomorphic. Is this itself a field in math? Do you know a book with good examples about it? Thank you! Nov 28, 2015 at 12:15
• @GniruT: It is indeed a powerful statement. One might define "abstract algebra" as the study of algebraic structures [such as groups] and isomorphisms between them (I think Birkhoff and Mac Lane give a similar definition in A Survey of Modern Algebra, towards the end of Ch. 1). So in a very general sense, yes, it is a field in itself. I'm not sure of books that focus solely on isomorphisms, but any book on group theory will cover the isomorphism theorems. The Fascination of Groups by F.J. Budden is not great as a standalone text, but has loads of examples. Nov 28, 2015 at 14:17
• @GniruT Practically always we investigate objects only up to isomorphism (preferably: up to natural isomorphism), whatever isomorphism means in that context. Also, the task of classifying objects according to their isomorphism class is classic. For sets: isomorphic means same cardinality, so cardinality is the "classifier". For vector spaces: isomorphic means same dimension, so dimension (i.e., cardinality of a base) is our classifier. I is a bit more complex but still not too difficult (you'll probably encounter it in your book sooner or later) to classify finite abelian groups. Nov 28, 2015 at 14:21
• @GniruT Then again, ignoring everything beyond isomorphism may be short-sighted. The set of isomorphisms between isomorphic objects has again an interesting structure. If - via one such isomorphism - we consider thius set as set of automorphisms of one object, interesting groups (and other algebraic structures) come up: For sets of $n$ elements, this gives us the permutation group $S_n$; for vector spaces of dimension $n$ over$\Bbb R$, say, this gives us the general linear group $GL(n,\Bbb R)$ of invertible linear maps (or matrices, if you prefer); and so on Nov 28, 2015 at 14:24
• @GniruT one other thing to bear in mind is that the fact that two groups are known to be isomorphic is very far from knowing how to compute such an isomorphism in an efficient way. It is for example known that the groups used for eliptic curves cryptography are all finite abelian and it is quite easy for any given one to figure out that it must be isomorphic to $\mathbb{Z}_a\times\mathbb{Z}_b$ for appropriate $a,b$. The whole premise of ECC is that it is hard to find discrete logarithms in those groups. Yet that is trivial in the isomorphic copies.
– DRF
Nov 28, 2015 at 21:12

In loose terms it means you can't tell them apart. They are the same except that the elements have different names. For example the group $Z_2 = \{0,1\}$ with the obvious rule for multiplication is isomorphic to the group {even, odd} with the usual rule for addition.

@Dorabell 's confusion (see his comment below) was my fault for calling the operation on $\{0,1\}$ "multiplication". "Addition mod $2$" would have been better. But his example is instructive in another way. The set $\{1,-1\}$ with the obvious multiplication is isomorphic too. That's interesting because the bijective homomorphism to $\{0,1\}$ maps $1$ to $0$ and $-1$ to $1$. You can't know what "$1$" means without the context.

• Surely you mean $Z_2 = \{-1, 1\}$? I can't think of any "obvious" rule for multiplication that makes $\{0, 1\}$ into a group! Nov 28, 2015 at 8:18
• @Dorebell: The obvious rule is, obviously, $0\cdot 1=1\cdot 0=1$ and $1\cdot 1=0\cdot 0=0$. Nov 28, 2015 at 11:35
• @tomasz: I see your point, and I know why you called it "multiplication", but I'd personally refer to that as an obvious rule for addition. Nov 28, 2015 at 17:10
• @Ethan Bolker: I disagree with the statement that "you can't tell them apart", Actually it is the opposite! Many time two groups look very different and turn out to be isomorphic. (Positive reals under multiplication on oneside and all real numbers under addition on the other side) Dec 1, 2020 at 9:07
• @PVanchinathan You can't tell them apart as groups. If all you know is the multiplication table with uninformative names for the elements they look the same. Dec 1, 2020 at 11:57

When we say that two groups are isomorphic, we are saying that they have the same structure and invariants as groups. An isomorphism between two groups do more than matching elements: it matches subgroups, normal subgroups, characteristic subgroups, conjugacy classes, $p$-subgroups, Frattini groups, ...

In other words, two isomorphic groups can be considered as the same object in the category of all groups. I don't know if it answers your question.

• – lhf
Nov 27, 2015 at 19:16
• @lhf I didn't know it, thank you. Nov 27, 2015 at 20:02

It means that, even though the groups contain different elements and combine according to different rules, they are nevertheless from the perspective of group theory essentially identical in every important respect.

The easiest example of this comes from elementary school arithmetic: We learn at an early age that adding two odds gives an even, etc. We also learn (a couple of years later) that multiplying two negatives gives a positive, etc. In fact the rules for adding evens and odds are exactly the same as the rules for multiplying positives and negatives.

More precisely, if you swap the word "positive" for "even", "negative" for "odd", and "add" for "multiply", then any true sentence you can write about addition of even and odd numbers becomes a true sentence about multiplying positive and negative numbers, and vice versa. Seen from an abstract level, they are groups with exactly the same structure.

They are, of course, different groups: Addition and multiplication are not the same operation, positive numbers do not need to be even, and so forth. Isomorphism describes the equivalence that you notice when you see past the particular details and focus on the structural relationships among the parts.

• Ah, as in the 'probabilistic way/mode of thinking' here, I take it? (In the sense that you use the notion to abstract away or put into a black box the labels/names, which are arbitrary anyway, so that you identify objects that are alike in the respects you care about.) Great way to articulate this; I feel happy when I think of or stumble on one. Nov 28, 2015 at 6:35

As other answers already point out, isomorphism is just relabeling elements and renaming operation, but all relations you could think of such as subgroups, quotient groups, generators etc. are preserved, up to relabeling.

Let me try to draw an analogy with similar but undoubtedly more familiar concept: what does it mean that two triangles (in a plane) are congruent? Well, let's say that we have two congruent triangles $\triangle ABC\cong\triangle DEF$. What it means is that we can move one triangle without deformation and overlap it with the other (more precisely, there is an isometry between them). So, how are these two triangles related? Well, for all intents and purposes, these are the same triangles, just with relabeled vertices and differently positioned in space, but with sides of the same length, angles of the same measure, radii of circumscribed circle of the same length, equally positioned orthocenter with respect to vertices... i.e. anything we would like to know about a triangle. When someone says: "Draw a triangle with these and these side lengths.", we don't ask where, or what should we name vertices, we just draw it. Because it's not important: we care about triangles up to congruence. Just as we care about groups up to isomorphism.

• I disagree with the first statement. It is not relabelling of elements: that applies only to AUTOMORPHISMS. Your elaboration with triangles is good Dec 1, 2020 at 9:03
• @P Vanchinathan, we must be using the word relabel differently. To me it means: apply a new label to. Dec 1, 2020 at 9:20
• Please see my answer which I have posted just now. Dec 1, 2020 at 9:32

I think that the point is to settle the definition of "isomorphic groups" on another basis than the one invoking the isomorphism (which sounds like swapping the effect with the cause), and then to show that the "operation-preserving property" of (necessarily) a bijection between the two groups is indeed the enabling property to allow the two groups to be isomorphic (according to the said different definition). Hereafter is an attempt to do so.

A group $$G$$ shows its structure as soon as we allow the internal operation to fully deploy its effects. Accordingly, we can reasonably state the following:

Definition 1. The structure of a group $$G$$ is the set $$\theta_G:=\{\theta_a, a\in G\}\subseteq \operatorname{Sym}(G)$$, where $$\theta_a$$ is the bijection on $$G$$ defined by: $$g\mapsto \theta_a(g):=ag$$.

Here a problem arises, if we want to determine whether two groups, $$G$$ and $$\tilde G$$, "have the same structure", since in general $$\operatorname{Sym}(G)\cap\operatorname{Sym}(\tilde G)=\emptyset$$, and then any attempt to "compare by overlapping" the structures $$\theta_G$$ and $$\tilde\theta_\tilde G$$ is doomed to fail. We can overcome this issue "by transporting" the structure of $$G$$ in $$\operatorname{Sym}(\tilde G)$$, and see whether we can make the "transported $$\theta_G$$" to overlap with $$\tilde\theta_{\tilde G}$$. If we succeed, then we can rightly say that $$G$$ and $$\tilde G$$ are isomorphic, since we have been able to bring the structure of one onto precisely that of the other. So, with reference to the following diagram:


we set forth this other:

Definition 2. Two groups, $$G$$ and $$\tilde G$$, are isomorphic if there is a bijection $$f\colon G\to \tilde G$$ such that the above diagram commutes, namely:

$$\tilde\theta=\varphi^{(f)}\theta f^{-1}\tag 1$$

where $$\varphi^{(f)}\colon \operatorname{Sym}(G)\to \operatorname{Sym}(\tilde G)$$ is the "structure transporting" bijection defined by $$\sigma\mapsto f\sigma f^{-1}$$.

Now, as a caracterization of such an "enabling" bijection $$f$$, the following holds:

Claim. Two groups $$G$$ and $$\tilde G$$ are isomorphic (as per the Definition 2) if and only if there is a bijection $$\psi\colon \tilde G \to G$$ such that:

$$\psi(\tilde a\tilde g)=\psi(\tilde a)\psi(\tilde g), \space\space\forall \tilde a,\tilde g\in \tilde G \tag 2$$

Proof.

\begin{alignat}{1} &\tilde\theta=\varphi^{(f)}\theta f^{-1} &\iff \\ &\tilde\theta_\tilde a(\tilde g)=(\varphi^{(f)}\theta f^{-1})(\tilde a)(\tilde g), \space\space\forall \tilde a,\tilde g\in \tilde G &\iff \\ &\tilde a\tilde g=(\varphi^{(f)}\theta f^{-1})(\tilde a)(\tilde g), \space\space\forall \tilde a,\tilde g\in \tilde G &\iff \\ &\tilde a\tilde g=(\varphi^{(f)}(\theta_{f^{-1}(\tilde a)}))(\tilde g), \space\space\forall \tilde a,\tilde g\in \tilde G &\iff \\ &\tilde a\tilde g=(f\theta_{f^{-1}(\tilde a)}f^{-1})(\tilde g), \space\space\forall \tilde a,\tilde g\in \tilde G &\iff \\ &\tilde a\tilde g=f(\theta_{f^{-1}(\tilde a)}(f^{-1}(\tilde g)), \space\space\forall \tilde a,\tilde g\in \tilde G &\iff \\ &\tilde a\tilde g=f(f^{-1}(\tilde a)f^{-1}(\tilde g)), \space\space\forall \tilde a,\tilde g\in \tilde G &\iff \\ &f^{-1}(\tilde a\tilde g)=f^{-1}(\tilde a)f^{-1}(\tilde g), \space\space\forall \tilde a,\tilde g\in \tilde G\\ \tag 3 \end{alignat}

So, $$(1)\Longrightarrow (2)$$, by setting $$\psi:=f^{-1}$$, and $$(2)\Longrightarrow (1)$$, by setting $$f:=\psi^{-1}$$. $$\space\space\Box$$

Therefore, a bijection between two groups with the property $$(2)$$ is rightly called isomorphism.

As I have commented below the answers Bamboo and Ethan Bolker, ismomorphism between two groups does not mean "they are one and the same". They are same only theoretically, i.e., algebraically, but not practically. Actually they could be differnet practically is the reason isomorphisms are useful!

Isomorphism only means what it says, a homomorphism which is bijective. As a consequence two isomorphic groups share many properties, number of elements of a specific order, being abelian, being solvable/nilpotent etc.

Actually isomorphism means two vastly different groups can be understood easily in terms of each other. For example in those days when there were no calculators (for get computers) one used logarithm tables to carry out multiplication. The reason that adding the logarithms and then finding the antilogarithm produces the product of two numbers is because: logarithm gives an isomorphism of the group of positive real numbers under multiplication with the group of all real numbers under addition.

Knowing a newly found group is isomorphic to a better understood known group gives us a handle to work with the former.

Many theorems are celebrated because they show two groups occurring in different contexts are isomorphic. (de Rham's theorem in differential geometry/topology, Artin's Reciprocity Law in Algebraic Number Theory)

Easier example: All polynomials with integer coefficients in a single variable is a group under usual addition of polynomials and is isomorphic to the group of all POSITIVE RATIONAL numbers under usual multiplication of numbers.

We'll try to build up to the definition of group isomorphism.

Suppose $$(G, \cdot)$$ is a group, and say we have a bijection $$\Phi : G \rightarrow \Phi(G)$$ ( which can be thought of as a relabelling $$g \mapsto \Phi(g)$$ of elements of $$G$$ ). A natural question arises : Can we endow $$\Phi(G)$$ with a group operation $$\ast$$ such that groups $$(G, \cdot)$$ and $$(\Phi(G), \ast)$$ "become copies of one another", i.e. are such that $$( x \cdot y = z \text{ in group } G ) \iff ( \Phi(x) \ast \Phi(y) = \Phi(z) \text{ in group } \Phi(G) )$$ ?

It's clear $$\Phi(x) \ast \Phi(y) := \Phi(x \cdot y)$$ is the only potential candidate for the definition of $$\ast$$ , and a simple check shows this definition actually works : It makes $$(\Phi(G), \ast)$$ a group and also a copy of $$(G, \cdot)$$ in the above sense.

This makes us arrive at the formal definition : Groups $$(G, \cdot)$$ and $$(G', \ast)$$ are called isomorphic if there is a bijection $$\phi : G \rightarrow G'$$ such that $$\phi(x \cdot y) = \phi(x) \ast \phi(y)$$.